Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 23: Drawing The Nyquist Plot
Overview In this Lecture, you will learn: Review of Nyquist Drawing the Nyquist Plot Using the Bode Plot What happens at r = Poles on the imaginary axis Phase Margin and Gain Margin Reading Stability Margins off the Nyquist Plot M. Peet Lecture 23: Control Systems 2 / 28
Review Systems in Feedback The closed loop is We want to know when kg(s) 1 + kg(s) 1 + kg(s) = Question: Does 1 k + G(s) have any zeros in the RHP? - u(s) + k G(s) y(s) M. Peet Lecture 23: Control Systems 3 / 28
Review The Nyquist Contour Definition 1. The Nyquist Contour, C N is a contour which contains the imaginary axis and encloses the right half-place. The Nyquist contour is clockwise. A Clockwise Curve Starts at the origin. Travels along imaginary axis till r =. At r =, loops around clockwise. Returns to the origin along imaginary axis. r = We want to know if 1 k + G(s) has any zeros in the Nyquist Contour M. Peet Lecture 23: Control Systems 4 / 28
Review Contour Mapping Principle Key Point: For a point on the mapped contour, s = G(s), s = G(s) We measure θ, not phase. To measure the 36 resets in G(s) We count the number of +36 resets in θ! We count the number of times C G encircles the origin Clockwise. The number of clockwise encirclements of is The # zeros # poles in the RHP s s * = G(s) θ = < G(s) M. Peet Lecture 23: Control Systems 5 / 28
The Nyquist Contour Closed Loop The number of unstable closed-loop poles is N + P, where N is the number of clockwise encirclements of 1 k. P is the number of unstable open-loop poles. If we get our data from Bode, typically P = -1/k How to Plot the Nyquist Curve? M. Peet Lecture 23: Control Systems 6 / 28
Plotting the Example How are we to plot the Nyquist diagram for G(s) = τ 1 = 1 τ 2 = 1 1 First lets take a look at the root locus. 1 (τ 1 s + 1)(τ 2 s + 1) Obviously stable for any k >. 6 Root Locus 4 2 2 4 6 12 1 8 6 4 2 2 M. Peet Lecture 23: Control Systems 7 / 28
The Nyquist Plot Bode Plot: Lets look at the Frequency Response. The Bode plot can give us information on G at different frequencies. Point ω G G A.1 1 B 1 45.7 C 3 9.3 D 1 135.7 E 1 175.1 Magnitude (db) 2-2 -4-6 -8-1 -12-45 A A Bode Diagram B C B D E The last two columns give us points on the Nyquist diagram. Phase (deg) -9-135 C D E -18 1-2 1-1 1 1 1 1 2 1 3 Frequency (rad/sec) M. Peet Lecture 23: Control Systems 8 / 28
The Nyquist Plot Plot the points from the Bode Diagram. Point ω G G A.1 1 B 1 45.7 C 3 9.3 D 1 135.7 E 1 175.1 We get the upper half of the Nyquist diagram from symmetry..8.6.4.2 -.2 D E C A -.4 -.6 B -.8 -.2.2.4.6.8 1 1.2 M. Peet Lecture 23: Control Systems 9 / 28
The Nyquist Plot.8.6.4.2 D E A -.2 C -.4 -.6 B -.8 -.2.2.4.6.8 1 1.2 There are no encirclements of 1 k. Stable for all k >. We already knew that from Root Locus. M. Peet Lecture 23: Control Systems 1 / 28
The Nyquist Plot Example 2 G(s) = First lets take a look at the root locus. 1 (s + 1) 3 We expect instability for large k. 2 Root Locus 1.5 1.5.5 1 1.5 2 2.5 2 1.5 1.5.5 M. Peet Lecture 23: Control Systems 11 / 28
The Nyquist Plot Bode Plot: Lets look at the Frequency Response. The Bode plot can give us information on G at different frequencies. Point ω G G A.1 1 B.28 45.95 C 1 135.35 D 1.8 18.1 E 1 26.1 Magnitude (db) 1-1 -2-3 -4-5 -6-7 -45 A A Bode Diagram B B C D E Phase (deg) -9-135 -18 C D -225 E -27 1-2 1-1 1 1 1 Frequency (rad/sec) M. Peet Lecture 23: Control Systems 12 / 28
The Nyquist Plot Plot the points from the Bode Diagram. Point ω G G A.1 1 B.28 45.95 C 1 135.35 D 1.8 18.1 E 1 26.1.8.6.4.2 D E A Point D is especially important. -.2 C -.4 -.6 B -.8 -.4 -.2.2.4.6.8 1 1.2 M. Peet Lecture 23: Control Systems 13 / 28
The Nyquist Plot.8.6.4.2 D E A -.2 C -.4 -.6 B -.8 -.4 -.2.2.4.6.8 1 1.2 Point D: Two CW encirclements when 1 k <.1 (N=2). Instability for 1 k >.1 Stable for k < 1. Could have used Routh Table. M. Peet Lecture 23: Control Systems 14 / 28
The Nyquist Plot Another Problem: Recall the non-inverted pendulum with PD feedback. G(s) = s + 1 s 2 + g l Magnitude goes to at ω = g l. Question How do we plot the Nyquist Diagram? 16 14 12 1 Bode Diagram 6 Magnitude (db) 8 6 4 2 4 2 2 4 45 2 Phase (deg) 45 9 4 6 135 1 2 1 1 1 1 1 1 2 Frequency (rad/sec) 2 1.5 1.5.5 1 1.5 2 2.5 M. Peet Lecture 23: Control Systems 15 / 28
The Nyquist Plot Problem: The Nyquist Contour passes through a pole. Because of the pole, the argument principle is invalid. What to do? r = We Modify the Nyquist Contour. We detour around the poles. Can detour to the right or left. If we detour to the left, then the poles count as unstable open loop poles. P=2 Assume we detour to the right. P= M. Peet Lecture 23: Control Systems 16 / 28
The Nyquist Plot Look at the detours at small radius. Obviously, magnitude Before the Detour, the phase from the pole is (s p) = 9 < (s - p) = 8 o In the middle of the Detour, the phase from the pole is (s p) = At the end of the Detour, the phase from the pole is (s p) = 9 The total phase change through the detour is 18. Corresponds to a CW loop at large radius. < (s - p) = -5 o If there are two or more poles, there is a -18 loop for each pole. M. Peet Lecture 23: Control Systems 17 / 28
The Nyquist Plot Look at the following example: G(s) = s + 2 s 2 There are 2 poles at the origin. At ω =, G() = 18 G() = 2 poles means 36 loop at ω = M. Peet Lecture 23: Control Systems 18 / 28
The Nyquist Plot Lets re-examine the pendulum problem with derivative feedback. ω= (g/l) 1 5 5 1 1 5 5 1 Now we can figure out what goes on at. There is a 18 loop at each ω = g l. M. Peet Lecture 23: Control Systems 19 / 28
The Nyquist Plot Conclusion: The loops connect in a non-obvious way! 15 ω= (g/l) 1 1 5 5 5 5 1 1 1 5 5 1 15 4 2 2 4 6 8 1 12 14 For < 1 k < 1, we have N = 1 M. Peet Lecture 23: Control Systems 2 / 28
Stability Margins Recall the definitions of Gain Margin. Definition 2. The Gain Margin, K m = 1/ G(ıω) when G(ıω) = 18 K m is the maximum stable gain in closed loop. It is easy to find the maximum stable gain from the Nyquist Plot. M. Peet Lecture 23: Control Systems 21 / 28
Stability Margins Example Recall G(s) = 1 (s + 1) 3.8.6 Stability: Stable for k < 1..4.2 K m = 1 or 2dB. D E A -.2 C -.4 -.6 B -.8 -.4 -.2.2.4.6.8 1 1.2 M. Peet Lecture 23: Control Systems 22 / 28
Stability Margins Example Suspension System with integral feedback There is a pole at the origin. CW loop at. 8 5 6 4 3 2 4 } 9.5 1 1 2 2 3-2 4 5-4 2 2 4 6 8 1 Conclusion: Stable for 1 k Stable for k <.15 > 9.5. K m =.15 or 19.5dB -6-8 -14-12 -1-8 -6-4 -2 2 M. Peet Lecture 23: Control Systems 23 / 28
Stability Margins Question: What is the effect of a phase change on the. A shift in phase changes the angle of all points. A Rotation about the origin. Will we rotate into instability?.8.8.6.6.4.4.2.2 -.2 -.2 -.4 -.4 -.6 -.6 -.8 -.2.2.4.6.8 1 1.2 -.8 -.2.2.4.6.8 1 1.2 M. Peet Lecture 23: Control Systems 24 / 28
Stability Margins Recall the definitions of Phase Margin. 3 Definition 3. The Phase Margin, Φ M is the uniform phase change required to destabilize the system under unitary feedback. 2 1 1 Φ M 2 3 3 2 1 1 2 3 3 2 1 1 2 3 3 2 1 1 2 3 M. Peet Lecture 23: Control Systems 25 / 28
Stability Margins Example The Suspension Problem 2 1.5 1.5 Φ M -.5-1 -1.5-2 -3-2.5-2 -1.5-1 -.5.5 1 Looking at the intersection with the circle: Phase Margin: Φ M = 4 Gain Margin is infinite. M. Peet Lecture 23: Control Systems 26 / 28
Stability Margins Example The Inverted Pendulum Problem 1.5 1.5 Φ M -.5-1 -1.5-2 -1.5-1 -.5.5 1 Even though open-loop is unstable, we can still find the phase margin: Phase Margin: Φ M = 35 Gain Margin is technically undefined because open loop is unstable. There is a minimum gain, not a maximum. M. Peet Lecture 23: Control Systems 27 / 28
Summary What have we learned today? Review of Nyquist Drawing the Nyquist Plot Using the Bode Plot What happens at r = Poles on the imaginary axis Phase Margin and Gain Margin Reading Stability Margins off the Nyquist Plot Next Lecture: Controller Design in the Frequency Domain M. Peet Lecture 23: Control Systems 28 / 28